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$[~1~]\qquad \D \sum_{n=1}^\infty \frac{x^{n+1}}{n-\alpha} =\sum_{n=1}^\infty (-1)^{n-1}\L(\frac{1}{n}+\frac{1-x}{n-\alpha}\R)\frac{(1-\alpha)_nn!}{(1-\alpha)_{2n}}\L(\frac{x^2}{1-x}\R)^n $

$[~2~]\qquad \D \sum_{n=1}^\infty \frac{2nx^n}{n^2-\alpha} =\sum_{n=1}^\infty \frac{(-1)^{n-1}(2n)!}{n}\L(\frac{x}{(1-x)^2}\R)^n\prod_{j=1}^{n}\frac{1}{j^2-\alpha} $

$[~3~]\qquad \D \sum_{n=1}^\infty \frac{x^n}{n^{2m+1}} =\sum_{n=1}^\infty \frac{(-1)^{n-1}{\zeta}^{\star}_{\le n}({\{2\}}^m)}{2n}\binom{2n}{n}\L(\frac{x}{(1-x)^2}\R)^n $

$[~4~]\qquad \D \sum_{n=1}^\infty \frac{21n-8}{n^3\binom{2n}{n}^3}=\zeta(2) $

$[~5~]\qquad \D \sum_{n=1}^\infty \frac{(-1)^{n-1}(56n^2-32n+5)}{4n^3(2n-1)^2}\frac{n!^3}{(3n)!}=\zeta(3) $

$[~6~]\qquad \D \sum_{n=1}^\infty\frac{1}{(n-\alpha)^3}=\sum_{n=1}^\infty \frac{(-1)^{n-1}(5n^2-4\alpha n+\alpha^2)}{2n^3(n-\alpha)^2}\frac{n!^2}{(1-\alpha)_{2n}} $

$[~7~]\qquad \D \sum_{n=m}^\infty \L(\frac{2(2m-1)}{m^3}-\frac{1}{(n+1)^2}\R)\frac{1}{\binom{n}{m}^2}=\frac{3}{m^2} $

$[~8~]\qquad \D \sum_{n=1}^\infty \frac{1}{(n-x)(n-y)}=\sum_{n=1}^\infty \frac{3n-x-y}{n\binom{2n}{n}}\frac{(1-x+y,1+x-y)_{n-1}}{(1-x,1-y)_n} $

$[~9~]\qquad \D A(k_1,\cdots,k_r)=\sum_{0< n_1<\cdots< n_r}\frac{1}{n_1^{k_1}\cdots n_r^{k_r}\binom{2n_r}{n_r}}$と定義すれば$\qquad\D \zeta(r+2)=A({\{1\}}^r,2)+2\sum_{j=0}^rA({\{1\}}^j,r-j+2)$

$[10]\qquad \D \sum_{n=1}^\infty \frac{1}{(n-x)(n-y)} =\sum_{n=1}^\infty \frac{(-1)^{n-1}(10n^2-(3+x+7y)n+y(1+x+y))}{n\binom{2n}{n}}\frac{(1-y)_{n-1}(1+x-y)_{2n-2}}{(1-x)_n(1-y)_{2n}} $

$[11]\qquad \D \sum_{n=1}^\infty \frac{1}{(n-x)(n-y)}=\sum_{n=1}^\infty \frac{(-1)^{n-1}(2n-x)}{n(n-x)}\frac{(1-x+y)_{n-1}}{(1-y)_n} $

$[12]\qquad \D \sum_{n=1}^\infty \frac{1}{(n-x)(n-y)} =\sum_{n=1}^\infty \frac{(21n-8)n^2-14(x+y)n^2+(x^2+y^2+11xy+4x+4y)n-(2+x+y)xy}{n\binom{2n}{n}}\frac{(1-x,1-y,1+x-y,1-x+y)_{n-1}}{(1-x,1-y)_{2n}} $

$[13]\qquad\D \sum_{n=1}^\infty \frac{1}{(n-x)(n-y)} =\sum_{n=1}^\infty \L(\frac{2n-x}{n-y}+\frac{2n^2+(1-4x+2y)n+x-y}{2n(2n-1)} \R)\frac{(-1)^{n-1}(1-x+y,1+x-y)_{n-1}}{(1-x)_{2n}} $

$[14]\qquad\D 2\zeta(2r)=(-1)^{r-1}\sum_{n=1}^\infty \frac{2n+1}{2n-1}\frac{(-1)^{n-1}{\zeta}_{< n}({\{2\}}^{r-1})}{n^2\binom{2n}{n}} +4\sum_{j=0}^{r-1}(-1)^j\sum_{n=1}^\infty \frac{(-1)^{n-1}{\zeta}_{< n}({\{2\}}^j)}{n^{2r-2j}\binom{2n}{n}} $

$[15]\qquad\D 2\zeta(2r+1)=(-1)^{r-1}\sum_{n=1}^\infty \frac{(-1)^{n-1}{\zeta}_{< n}({\{2\}}^{r-1})}{n^3\binom{2n}{n}} +4\sum_{j=0}^{r-1}(-1)^j\sum_{n=1}^\infty \frac{(-1)^{n-1}{\zeta}_{< n}({\{2\}}^{j})}{n^{2r-2j+1}\binom{2n}{n}} $

$[16]\qquad\BA\D \sum_{j=1}^r 2^{j}\sum_{\substack{k_1+\cdots+k_j=r\\k_i>0}}\zeta(2m+k_1,k_2,\cdots,k_{j-1},\overline{k_j})=-\sum_{0< n_0\le \cdots\le n_m}\frac{1}{n_0^2\cdots n_{m-1}^2 n_{m}^r}\frac{\binom{2n_m}{n_m}}{2^{2n_m}} \EA$

$[17]\qquad\D \sum_{n=1}^\infty \frac{(n-1)!^2}{(1-x,1+x)_{n}}=\sum_{n=1}^\infty (3n+x)\frac{(n-1)!(1+x)_{n-1}^2}{(1-x)_n(1+x)_{2n}} $

$[18]\qquad\D \sum_{n=1}^\infty \frac{(n-1)!^2}{(1-x,1+x)_{n}}=\sum_{n=1}^\infty (11n^3-17(1-x)n^2+(5-11x+7x^2)n+x(1-x)^2)\frac{(2n-2)!(1+x)_{n-1}(1+x)_{2n-2}}{(x,1-x)_n(1+x)_{3n}} $

$[19]\qquad\D \sum_{n=1}^\infty \frac{(n-1)!^2}{(1-x,1+x)_{n}}=\sum_{n=1}^\infty (46n^3-(34-7x)n^2+(1-x)(7+4x)n+x(1+x-x^2))\frac{(2n-2)!^2(1+x)_{n-1}^2}{(n-1)!(1-x)_{2n}(1+x)_{3n}} $

$[20]\qquad\D \zeta(r+2) =(-1)^r\sum_{n=1}^\infty \frac{(-1)^{n-1}{\zeta}_{< n}({\{1\}}^r)}{n^2\binom{2n}{n}}+\sum_{j=0}^r(-1)^j\sum_{n=1}^\infty \frac{(-1)^{n-1}{\zeta}_{< n}({\{1\}}^j)}{n\binom{2n}{n}}\L(\frac{1}{(2n)^{r-j+1}}+\frac{2}{(2n-1)^{r-j+1}}\R) $

$[21]\qquad\D \sum_{n=1}^\infty \frac{(n-1)!^2}{(1-x,1-y)_n}=\sum_{n=1}^\infty \frac{3n^2-3(x+y)n+x^2+xy+y^2}{(n-x)(n-y)}\frac{(1-x,1-y)_{n-1}}{(1-x-y)_{2n}} $

$[22]\qquad\D {\zeta}^{\star}({\{2\}}^{r})=(-1)^{r}A({\{2\}}^{r})+4\sum_{j=0}^{r-1}(-1)^jA({\{2\}}^{j},2r-2j) $

$[23]\qquad\D \zeta(2)=\sum_{n=0}^\infty \frac{(-1)^n2^{4n+2}}{(2n+1)^2\binom{2n}{n}\binom{4n+2}{2n+1}}-\frac{1}{3}\sum_{n=1}^\infty \frac{(-1)^{n-1}2^{4n-1}}{n^2\binom{2n}{n}\binom{4n}{2n}} $

$[24]\qquad\D 7\zeta(3)=\sum_{n=0}^\infty \frac{(-1)^n2^{4n+4}}{(2n+1)^3\binom{2n}{n}\binom{4n+2}{2n+1}}+\sum_{n=1}^\infty \frac{(-1)^{n-1}2^{4n-1}}{n^3\binom{2n}{n}\binom{4n}{2n}} $

$[25]\qquad\D \sum_{n=1}^\infty \frac{(1-x,1-y)_{n-1}}{n!^2}=\sum_{n=1}^\infty (10n^3-(3-7x-8y)n^2+(x^2+5xy-x-3y)n-(1-x)xy)\frac{(-1)^{n-1}(1+x)_{2n-2}(1+x,1-y,1+y)_{n-1}}{n!(2n)!(1+x+y)_{2n}} $

$[26]\qquad\D \sum_{n=1}^\infty \frac{(c,d)_{n-1}}{(a)_n(b+c+d)_{n-1}} =\sum_{n=1}^\infty (3n^2+(-6+2a+3b+2c+2d)n+(a+b-1)(b+c+d-2)+(c-1)(d-1))\frac{(b+c,b+d,c,d)_{n-1}}{(a,a+b)_n(b+c+d)_{2n-1}} $

$[27]\qquad\D \sum_{n=1}^\infty \frac{(a+c,a+d)_{n-1}}{(a)_n(a+b+c+d)_{n-1}} =\sum_{n=1}^\infty p(n)\frac{(1-c,1-d,b+c,b+d)_{n-1}}{(a,a+b+c+d)_n(b)_{2n}}\\ \qquad\qquad\qquad\qquad p(n)=((a+b+c+d+n-1)(b+2n-2)+(a+c)(a+d))(b+2n-1)+(a+c+d-n)(b+c+n-1)(b+d+n-1) $

$[28]\qquad\D \sum_{n=1}^\infty \frac{1}{(n-x)(n-y)} =\sum_{n=1}^\infty \left(\frac{(1-y)_{kn}}{n(1-y)_{(k+1)n}}+\sum_{m=1}^k\frac{(1-y)_{kn-m}}{(kn-m+1-x)(1-y)_{(k+1)n-m}} \right)(1+x-y)_{n-1} $

$[29]\qquad\D \sum_{n=1}^\infty \frac{(1+x)_{n-1}}{n!}z^{n-1} =\sum_{n=1}^\infty \frac{(-1)^{n-1}((2-z)n+zx)}{\binom{2n}{n}}\frac{(1-x,1+x)_{n-1}}{n!^2}z^{2n-2}(1-z)^{-n} $

$[30]\qquad\D \sum_{n=1}^\infty \frac{1}{n^2-x^2} =\sum_{n=1}^\infty \L( \frac{4n+1}{2(2n-1)}+\frac{3(2n-1)(3n-1)}{(2n-1)^2-x^2}+\frac{6n^2}{(2n)^2-x^2}\R)\frac{(-1)^{n-1}n!(1-x,1+x)_{n-1}}{(3n)!} $

$[31]\qquad\D \sum_{n=1}^\infty \frac{1}{n(n^2-x^2)} =\sum_{n=1}^\infty \L(\frac{1}{2n}+\frac{3n}{(2n)^2-x^2}+\frac{3(3n-1)}{(2n-1)^2-x^2}\R)\frac{(-1)^{n-1}n!(1-x,1+x)_{n-1}}{(3n)!} $

$[32]\qquad\D 8\beta(2r+2)=\sum_{j=0}^r (-1)^j2^{2r-2j+1}\sum_{n=1}^\infty \frac{4n-1}{n^2(2n-1)^{2r-2j+1}}\frac{2^{4n}\zeta_{< n}(\{2\}^j)}{\binom{2n}{n}\binom{4n}{2n}} -(-1)^r\sum_{n=1}^\infty \frac{2^{4n}\zeta_{< n}(\{2\}^r)}{n^2\binom{2n}{n}\binom{4n}{2n}} $

$[33]\qquad\D \sum_{n=1}^\infty \frac{1}{n^2(n^2-x)}=4\sum_{n=1}^\infty \frac{1}{n(n^2-x)}\sum_{m=1}^{n-1}\frac{m}{m^2-x} $

$[34]\qquad\D \zeta(2,3)=2\sum_{0< m< n}\frac{\binom{2m}{m}}{mn^4\binom{2n}{n}}+3\sum_{0< k< m< n}\frac{\binom{2m}{m}}{k^2mn^2\binom{2n}{n}}+12\sum_{0< k< m< n}\frac{\binom{2k}{k}}{km^2n^2\binom{2n}{n}} $

投稿日:20211230

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